题目
题型:不详难度:来源:
x2 |
a2 |
y2 |
b2 |
OF |
FP |
(1)设
1 |
2 |
| ||
2 |
OF |
FP |
(2)设|
OF |
3 |
4 |
OP |
答案
OF |
FP |
OF |
FP |
OF |
FP |
1 |
2 |
OF |
FP |
两式相除可得tanθ=2S,又
1 |
2 |
| ||
2 |
3 |
所以向量
OF |
FP |
(2)设P(x0,y0),F(c,0),所以
OF |
FP |
所以
OF |
FP |
1 |
c |
所以S=
1 |
2 |
3 |
4 |
3 |
2 |
所以|
OP |
|
(c+
|
由单调性可知当c=2时有最小值,此时x0=
5 |
2 |
(
|
(
|
10 |
所以椭圆方程为
x2 |
10 |
y2 |
6 |
核心考点
试题【如图,F为椭圆x2a2+y2b2=1(a>b>0)的右焦点,P为椭圆上一点,O为原点,记△OFP的面积为S,且OF•FP=1.(1)设12<S<32,求向量OF】;主要考察你对椭圆的几何性质等知识点的理解。[详细]
举一反三
x2 |
a2 |
y2 |
b2 |
a2-b2 |
c |
a |
x2 |
4 |
y2 |
m |
x2 |
m |
y2 |
9 |
x2 |
169 |
y2 |
144 |
x2 |
9 |
y2 |
16 |
x2 |
a2 |
y2 |
b2 |
a2 |
c |
3 |
A.
| B.
| C.
| D.
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